Question

Let $X=\{11,12,13,\dots .,40,41\}$ and $Y=\{61,62$, $63,\dots \dots ,90,91\}$ be the two sets of observations. If $\overline{x}$ and $\overline{y}$ are their respective means and ${\sigma }^2$ is the variance of all the observations in $X\cup Y$, then $\left|\overline{x}+\overline{y}-{\sigma }^2\right|$ is equal to

Answer 1

Answer: 603

$\overline{x}=\frac{\sum _{i=11}^{41}i}{31}=\frac{11+41}{2}=26(31\text{ elements })$
$\overline{y}=\frac{\sum _{j=61}^{91}j}{31}=\frac{61+91}{2}=76(31\text{ elements) }$
Combined mean, $\mu =\frac{31\times 26+31\times 76}{31+31}$
$=\frac{26+76}{2}=51$
${\sigma }^2=\frac{1}{62}\times \left(\sum _{i=1}^{31}{\left(x_i-\mu \right)}^2+\sum _{i=1}^{31}{\left(y_i-\mu \right)}^2\right)=705$
Since, $x_i\in X$ are in A.P. with 31 elements & common difference 1 , same is $y_i\in y$, when written in increasing order.
$\therefore \sum _{i=1}^{31}{\left(x_i-\mu \right)}^2=\sum _{i=1}^{31}{\left(y_i-\mu \right)}^2$
$=10^2+11^2+\dots ..+40^2$
$=\frac{40\times 41\times 81}{6}-\frac{9\times 10\times 19}{6}=21855$
$\therefore \left|\overline{x}+\overline{y}-{\sigma }^2\right|=|26+76-705|=603$